Problem 22
Question
Molecular shapes of \(\mathrm{SF}_{4}, \mathrm{CF}_{4}\) and \(\mathrm{XeF}_{4}\) are (a) the same, with 2,0 and 1 lone pairs of electrons respectively (b) the same, with 1,1 and 1 lone pairs of electrons respectively (c) different, with 0,1 and 2 lone pairs of electrons respectively (d) different, with 1,0 and 2 lone pairs of electrons respectively
Step-by-Step Solution
Verified Answer
(d) different, with 1, 0, and 2 lone pairs respectively.
1Step 1: Determine the Lewis Structure for SF4
To start, we need the Lewis structure for \(\mathrm{SF}_{4}\). Sulfur has 6 valence electrons and fluorine has 7 valence electrons. Therefore, \(\mathrm{SF}_{4}\) has a total of 34 valence electrons (6 from sulfur + 28 from four fluorines). Distributing these electrons, sulfur bonds with four fluorine atoms and has two remaining electrons, resulting in one lone pair on sulfur.
2Step 2: Determine Molecular Shape of SF4
\(\mathrm{SF}_{4}\) has 4 bonding pairs and 1 lone pair, which causes a seesaw shape due to the lone pair on the equatorial plane in a trigonal bipyramidal electronic geometry.
3Step 3: Determine the Lewis Structure for CF4
Next, construct the Lewis structure for \(\mathrm{CF}_{4}\). Carbon has 4 valence electrons, and each fluorine has 7 valence electrons, totaling 32 valence electrons for \(\mathrm{CF}_{4}\). Carbon forms four bonds with fluorine atoms, using all its available electrons for bonding and resulting in no lone pairs on carbon.
4Step 4: Determine Molecular Shape of CF4
\(\mathrm{CF}_{4}\) has 4 bonding pairs and 0 lone pairs, resulting in a tetrahedral shape as there are no lone pairs to distort it.
5Step 5: Determine the Lewis Structure for XeF4
Now, consider \(\mathrm{XeF}_{4}\). Xenon has 8 valence electrons, and each fluorine has 7, for a total of 36 valence electrons. Xenon forms four bonds with fluorine and uses two pairs of its electrons as lone pairs.
6Step 6: Determine Molecular Shape of XeF4
\(\mathrm{XeF}_{4}\) has 4 bonding pairs and 2 lone pairs. The electronic geometry is octahedral; however, the presence of two lone pairs makes the molecule adopt a square planar shape.
7Step 7: Compare Molecular Shapes and Lone Pairs
Finally, compare the results: \(\mathrm{SF}_{4}\) has 1 lone pair and a seesaw shape, \(\mathrm{CF}_{4}\) has no lone pairs with a tetrahedral shape, and \(\mathrm{XeF}_{4}\) has 2 lone pairs with a square planar shape. This shows the molecular shapes are different: \(\mathrm{SF}_{4}\) and \(\mathrm{XeF}_{4}\) each having different numbers of lone pairs and geometries.
Key Concepts
Lewis structureLone pairsElectronic geometry
Lewis structure
To determine molecular shapes, the Lewis structure is an essential starting point. A Lewis structure helps visualize how atoms connect in a molecule and shows the arrangement of electrons.
In any Lewis structure:
Understanding this structure is critical to predicting further molecular shape.
In any Lewis structure:
- Atoms are represented by their chemical symbols.
- Lines between atoms show bonds – each line stands for a shared pair of electrons.
- Dots around the atomic symbols represent lone pairs – non-bonded valence electrons.
Understanding this structure is critical to predicting further molecular shape.
Lone pairs
Lone pairs are valence electrons that do not participate in bonding. They are crucial in determining the molecular geometry because they occupy space around an atom. Although not visible in the molecule's "sticks" and "balls," lone pairs have an electrostatic impact on molecular shape.
In VSEPR theory (Valence Shell Electron Pair Repulsion theory):
Each lone pair affects the three-dimensional structure and directly impacts how the electrons distribute around the central atom.
In VSEPR theory (Valence Shell Electron Pair Repulsion theory):
- Lone pairs repel more strongly than bonding pairs, leading to adjustments in bond angles.
- These pairs influence the overall geometry due to their spatial properties.
Each lone pair affects the three-dimensional structure and directly impacts how the electrons distribute around the central atom.
Electronic geometry
Electronic geometry considers the spatial arrangement of all electron pairs, both bonding and lone pairs, around a molecule's central atom. This holistic view includes every region of negative electron density.
For instance, in \( \mathrm{SF}_4 \), the electronic geometry can be characterized as trigonal bipyramidal because it includes four bonds and one lone pair. However, when only shaping the visible molecular part, the lone pair modifies its structure into a seesaw form. \( \mathrm{CF}_4 \) maintains a tetrahedral electronic geometry, as there are no lone pairs to alter the symmetrical shape. Meanwhile, \( \mathrm{XeF}_4 \) features octahedral electronic geometry encompassing four bonds and two lone pairs.
Recognizing electronic geometry is key to predicting molecules' behaviors and interactions because it underpins why lone pairs result in their unique structures and orientations. By understanding these spatial and electronic influences, one can better understand molecular reactivity and properties.
For instance, in \( \mathrm{SF}_4 \), the electronic geometry can be characterized as trigonal bipyramidal because it includes four bonds and one lone pair. However, when only shaping the visible molecular part, the lone pair modifies its structure into a seesaw form. \( \mathrm{CF}_4 \) maintains a tetrahedral electronic geometry, as there are no lone pairs to alter the symmetrical shape. Meanwhile, \( \mathrm{XeF}_4 \) features octahedral electronic geometry encompassing four bonds and two lone pairs.
Recognizing electronic geometry is key to predicting molecules' behaviors and interactions because it underpins why lone pairs result in their unique structures and orientations. By understanding these spatial and electronic influences, one can better understand molecular reactivity and properties.
Other exercises in this chapter
Problem 21
Which one of the following molecules is expected to exhibit diamagnetic behaviour? [Main 2013] (a) \(\mathrm{C}_{2}\) (b) \(\mathrm{N}_{2}\) (c) \(\mathrm{O}_{2
View solution Problem 22
Which of the following compounds are covalent? (a) \(\mathrm{H}_{2}\) (b) \(\mathrm{CaO}\) (c) \(\mathrm{KCl}\) (d) \(\mathrm{Na}_{2} \mathrm{~S}\)
View solution Problem 22
Which of the following is the wrong statement (a) ONCl and \(\mathrm{ONO}^{-}\)are not isoelectronic. (b) \(\mathrm{O}_{3}\) molecule is bent (c) Ozone is viole
View solution Problem 23
Element \(X\) is strongly electropositive and element \(Y\) is strongly electronegative. Both are univalent. The compound formed would be [1980] (a) \(X^{+} Y^{
View solution